Convex Lens Real
Images:
Problem Solving

Six problems are presented with solutions
for your inspection. In each case we used the GUESS
formula in our solution.

Note that when we are solving these problems, we will use
Do instead of Do to represent object
distance, and Di instead of Di to
represent image distance. It is hoped that you will continue to
consider the o and i as subscripts and use them as
such in your own work. We will also use the exponent -1 instead of
writing out the inverse fraction in many cases. This is to save
key strokes!

Problem 1:

A convex lens with a focal length of 15 cm is placed 25 cm from
a lighted object. At what distance will an image be formed? Will
it be real?

Givens

f = 15 cm
Do = 25 cm

Unknown

Di = ?

Equation

1/f = 1/Do + 1/Di

Solving

1/Di = 1/f - 1/Do = 1/15 cm - 1/25 cm = 0.0267
cm-1

Di = (0.0267 cm-1)-1 =
37.5 cm

Because the object is further from the lens than the focal
point, the image will be real. Also notice that the
image distance is a positive one. This means that it is on the
other side of the lens and is therefore a real image. Finally,
the image distance is larger than the object distance.
Therefore the image is larger than the object. (Remember that
magnification is simply the ratio of image distance to object
distance, Hi/Ho.)

Problem 2:

An object is placed 25 cm from a convex lens which has a focal
length of 10 cm. At what distance will the image be formed? Will
it be real? Will it be enlarged or reduced in size?

Givens

f = 10 cm
Do = 25 cm

Unknown

Di = ?

Equation

1/f = 1/Do + 1/Di

Solving

1/Di = 1/f - 1/Do = 1/10 cm - 1/25 cm = 0.06
cm-1

Di = (0.06 cm-1)-1 =
16.67 cm

Because the object is further from the lens than the focal
point, the image will be real. Also notice that the
image distance is a positive one. This means that it is on the
other side of the lens and is therefore a real image. Finally,
the image distance is less than the object distance. Therefore
the image is smaller than the object. (Remember that
magnification is simply the ratio of image distance to object
distance, Hi/Ho.)

Problem 3:

Where should one place a lighted object so that the final image
is 4 meters away from a lens having a focal length of 20 cm? Will
the final image be enlarged or reduced?

Givens

f = 20 cm
Di = 400 cm

Unknown

Do = ?

Equation

1/f = 1/Do + 1/Di

Solving

1/Do = 1/f - 1/Di = 1/20 cm - 1/400 cm = 0.0475
cm-1

Di = (0.0475 cm-1)-1 =
21.05 cm

Because the image is further from the lens than the focal
point, the image will be real. The image distance is
much greater than the object distance. Therefore the image is
enlarged. (Remember that magnification is simply the ratio of
image distance to object distance, Hi/Ho.) This is
similar to having a slide projector. In fact, where wouldyou
place the slide you need to project? Yes, just a little ways
from the focal point of the lens.

Problem 4:

We have an object that's very far away (100 meters) compared to
the 15-cm focal length of our convex lens. Where will the image be
formed? What are its characteristics?

Givens

f = 15 cm
Do = 104 cm

Unknown

Di = ?

Equation

1/f = 1/Do + 1/Di

Solving

1/Di = 1/f - 1/Do = 1/15 cm - 1/104 cm =
0.06657 cm-1

Di = (0.066657 cm-1)-1 =
15.02 cm

Note that the image distance is very close to being the same
as the focal length. Objects that are very far away from convex
lenses will form images that are at or very close to the focal
point. This is helpful so that we can design cameras that
clearly focus objects at a wide range of distances, as long as
they are relatively far from the lens.

Because the object is further from the lens than the focal
point, the image will be real. The image distance is a
positive one which means that it is on the other side of the
lens and is therefore a real image. Finally, the image distance
is much less than the object distance. Therefore the image is
smuch maller than the object. (Remember that magnification is
simply the ratio of image distance to object distance,
Hi/Ho.)

Another way of thinking about this situation is to consider
that an object very far away presents the lens with light rays
that are almost parallel to one another. (The angle they make
when striking the lens is very small.) Therefore the place they
would be brought together must be close to the focal point.

Problem 5:

We have a convex lens with a focal length of 15 cm. We wish to
place an object so that the final image will be 1/3 the size of
the object. Where should we place that object?

Givens

f = 15 cm
Hi = 1/3 Ho

Unknown

Do = ?

Equations

Hi/Ho = Di/Do

1/f = 1/Do + 1/Di

Solving

Hi / Ho = (1/3 Ho) / Ho = 1/3 = Di / Do

Therefore Di = 1/3 Do

1/f = 1/Do + 1/Di = 1 / Do + 1 / (1/3 Do) = 1/Do + 3/Do =
4/Do

4/Do = 1/f ..... Do = 4 f = 4 x 15 cm =
60 cm

Proof: 1/Di = 1/f - 1/Do = 1/15 cm - 1/60 cm = 0.05
cm-1

Di = (0.05 cm-1)-1 = 20 cm

Hi/Ho = Di/Do = 20 cm / 60 cm = 1/3 QED

Problem 6:

We place an object 20 cm from a convex lens with a focal length
of 25 cm. Where is the image formed?

Givens

f = 25 cm
Do = 20 cm

Unknown

Di = ?

Equation

1/f = 1/Do + 1/Di

Solving

1/Di = 1/f - 1/Do = 1/25 cm - 1/20 cm = -0.01
cm-1

Di = (-0.01 cm-1)-1 =
-100 cm

We did get an answer. We worked carefully, but the answer
turned out negative. At this point we may not have a way to
think about negative answers, but with a non-positive answer we
conclude that the image is likely not real. To learn more about
what this negative answer means, go to the section on VIRTUAL
IMAGES.

We hope this set of problems is useful in helping you with your
problem-solving using convex lenses. Remember to use the GUESS
formula in all of your work. Include units in all of your work,
too, and be sure to check your final answers.

GUESS FORMULA:

G = Givens - write them all down with
symbols, numbers and units

U = Unknown - write down the unknown with a
question mark

E = Equation - write down the equation or
equations that apply in the problem
solve
the equation(s) for the desired unknown

S = Substitute - substitute the given
information into the solved equation

S = Solve - solve numerically, paying
attention to units
identify
the final answer by circling, boxing, or underlining it
clearly
check
the final answer to be sure it makes sense